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How much work was done by friction on the person?

  1. Feb 16, 2009 #1
    1. The problem statement, all variables and given/known data
    A person's mass is 28 kg. They climb the 4.8m ladder of a slide and reach a velocity of 3.2 m/s at the bottom of the slide. How much work was done by friction on the person?


    2. Relevant equations
    KE=1/2mv^2
    PE=mgh
    W=Fd
    i think these are all the equations that apply

    3. The attempt at a solution
    i think you could find the potential energy=28(9.8)(4.8), which equals the kinetic energy,
    then i dont know what to do but i know what i did is wrong
    thank you for any help
     
  2. jcsd
  3. Feb 16, 2009 #2
    Re: sliding(work)

    Use the work energy principle,

    [tex] \Delta E_k=W [/tex]

    What forces are doing work on the person?
     
    Last edited: Feb 16, 2009
  4. Feb 16, 2009 #3
    Re: sliding(work)

    i think you would have to solve for PE first but im not sure
     
  5. Feb 16, 2009 #4
    Re: sliding(work)

    Sure, that is a good first step.

    How about this, IF there was no friction, how fast would you expect the person to be going at the bottom of the slide?
     
  6. Feb 16, 2009 #5
    Re: sliding(work)

    she would be going about 10 m/s without friction
     
  7. Feb 16, 2009 #6
    Re: sliding(work)

    Okay, so then how much kinetic energy would she have if there was no friction? (v=9.7m/s)

    How much kinetic energy does she actually end up with?

    What happened to the energy?
     
    Last edited: Feb 16, 2009
  8. Feb 16, 2009 #7
    Re: sliding(work)

    she would end up with 1300J of kinetic energy when v = 9.7 m/s

    she would have 140 J of kinetic energy with the given 3.2 m/s
     
  9. Feb 16, 2009 #8
    Re: sliding(work)

    So if you consider that only gravity and friction do work on the person, what can you say.

    Remember the conservation of energy.
     
  10. Feb 16, 2009 #9
    Re: sliding(work)

    you could say that the gravitational potential energy is 140 J since PE=KE
     
  11. Feb 16, 2009 #10
    Re: sliding(work)

    No, that's not right. The gravitational potential energy at the top is mgh=1300J

    The kinetic energy at the bottom is 140J

    The difference is 1300J-140J = 1160J

    Energy cannot be destroyed, so somehow 1160J of energy was lost by the person.

    We haven't yet considered friction though, as the person slides down the slide friction does negative work on the person, slowing them down.
     
  12. Feb 16, 2009 #11
    Re: sliding(work)

    oh okay, sorry, you're right
    it makes more sense now thanks for your help again
     
  13. Feb 16, 2009 #12
    Re: sliding(work)

    np, don't be sorry! I just wanted you to understand that the only two forces acting on the person are the gravitational force which does work in the amount mgh=1300J, and the frictional force.

    Since we know [tex] \Delta E_k=W [/tex], we have [tex]1/2mv_2^2-1/2mv_1^2=1300J - W_f [/tex] where [tex]W_f[/tex] is the work done by friction. It is negative because it acts opposite to the motion of the person :)
     
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